Class 12: Application of Calculus – ISC Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{A company sells hand towels at Rs } 100 \text{ per unit. The fixed cost for the }$
$\displaystyle \text{company to manufacture hand towels is Rs 35000 and variable cost is estimated to be}$
$\displaystyle \text{ } 30\% \text{ of} \text{total revenue. What will be the total cost function for } \text{manufacturing hand}$
$\displaystyle \text{ towels? ISC 2024}$
$\displaystyle \text{(a) } 35000+3x \quad \text{(b) } 35000+30x$
$\displaystyle \text{(c) } 35000+100x \quad \text{(d) } 35000+10x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Let the number of units of manufacturing hand towels be }x$
$\displaystyle \text{and selling price per unit}=\text{Rs }100$
$\displaystyle \therefore \text{Revenue generated on selling }x\text{ units is}$
$\displaystyle R(x)=100x$
$\displaystyle \text{Variable cost}=30\%\text{ of total revenue}$
$\displaystyle =\frac{30}{100}R(x)=\frac{30}{100}\times100x=30x$
$\displaystyle \therefore \text{Total cost}=\text{Fixed cost}+\text{Variable cost}$
$\displaystyle =35000+30x$
$\displaystyle \therefore \text{Cost function for manufacturing hand towels is }C(x)=35000+30x$
$\\$

$\displaystyle \textbf{Question 2: } \text{If the total cost function is given by } C=x+2x^{3}-\frac{7}{2}x^{2}, \text{ find the}$
$\displaystyle \text{Marginal Average Cost function (MAC). ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, total cost function is}$
$\displaystyle C(x)=x+2x^{3}-\frac{7}{2}x^{2}$
$\displaystyle \text{Average cost is}$
$\displaystyle AC=\frac{C(x)}{x}=\frac{x+2x^{3}-\frac{7}{2}x^{2}}{x}$
$\displaystyle =1+2x^{2}-\frac{7}{2}x$
$\displaystyle \text{Marginal Average Cost function is}$
$\displaystyle MAC=\frac{d}{dx}(AC)$
$\displaystyle =\frac{d}{dx}\left(1+2x^{2}-\frac{7}{2}x\right)$
$\displaystyle =0+4x-\frac{7}{2}$
$\displaystyle \therefore MAC=4x-\frac{7}{2}$
$\\$

$\displaystyle \textbf{Question 3: } \text{The manufacturer of a pen fixes its selling price at Rs } 45, \text{ and the cost}$
$\displaystyle \text{function is } C(x)=30x+240. \text{ The manufacturer will begin to earn profit if he sells}$
$\displaystyle \text{more than } 16 \text{ pens. Why? Give one reason. ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Selling price of pen fixed at Rs }45$
$\displaystyle \text{The cost function is }C(x)=30x+240$
$\displaystyle \text{and the revenue function is }R(x)=45x$
$\displaystyle \text{So, the profit is}$
$\displaystyle P(x)=R(x)-C(x)=45x-(30x+240)$
$\displaystyle =45x-30x-240=15x-240$
$\displaystyle \text{For break-even point, we put }P(x)=0$
$\displaystyle P(x)=0\Rightarrow 15x-240=0$
$\displaystyle \Rightarrow 15x=240\Rightarrow x=\frac{240}{15}=16$
$\displaystyle \text{Since, the break-even point is }16.$
$\displaystyle \text{Therefore, the manufacturer will begin to earn profit, if}$
$\displaystyle \text{he sells more than }16\text{ pens.}$
$\\$

$\displaystyle \textbf{Question 4: } \text{The average cost function associated with producing and marketing } x \text{ units}$
$\displaystyle \text{of an item is given by } AC=x+5+\frac{36}{x}. \text{ (i) Find the total cost function.}$
$\displaystyle \text{(ii) Find the range of values of } x \text{ for which average cost is increasing. ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, average cost function is}$
$\displaystyle AC=x+5+\frac{36}{x}$
$\displaystyle \text{(i) We know that}$
$\displaystyle \text{Average cost}=\frac{\text{Total cost}}{x}$
$\displaystyle \Rightarrow AC=\frac{C(x)}{x}$
$\displaystyle \Rightarrow x+5+\frac{36}{x}=\frac{C(x)}{x}$
$\displaystyle \Rightarrow C(x)=x\left(x+5+\frac{36}{x}\right)$
$\displaystyle =x^{2}+5x+36$
$\displaystyle \text{Thus, the total cost function is }x^{2}+5x+36.$
$\displaystyle \text{(ii) Since, }AC=x+5+\frac{36}{x}$
$\displaystyle \frac{d}{dx}(AC)=\frac{d}{dx}\left(x+5+\frac{36}{x}\right)$
$\displaystyle =1+0-\frac{36}{x^{2}}$
$\displaystyle =1-\frac{36}{x^{2}}$
$\displaystyle \text{For AC to be increasing,}$
$\displaystyle \frac{d}{dx}(AC)>0$
$\displaystyle \Rightarrow 1-\frac{36}{x^{2}}>0$
$\displaystyle \Rightarrow x^{2}-36>0$
$\displaystyle \Rightarrow (x-6)(x+6)>0$
$\displaystyle \Rightarrow x<-6,\;x>6$
$\displaystyle \text{Hence, the average cost function increases, if the value of }x>6$
$\displaystyle \text{i.e. }(6,\infty)$
$\\$

$\displaystyle \textbf{Question 5: } \text{A monopolist's demand function is } x=60-\frac{p}{5}. \text{ At what level of output}$
$\displaystyle \text{will marginal revenue be zero? ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given monopolist's demand function is}$
$\displaystyle x=60-\frac{p}{5}$
$\displaystyle \Rightarrow 5x=300-p$
$\displaystyle \Rightarrow p=300-5x$
$\displaystyle \text{Thus, the revenue function is}$
$\displaystyle R(x)=p\cdot x$
$\displaystyle =(300-5x)x=300x-5x^{2}$
$\displaystyle \text{Now, marginal revenue }=\frac{dR}{dx}$
$\displaystyle =\frac{d}{dx}(300x-5x^{2})$
$\displaystyle =300-10x$
$\displaystyle \text{Then, the marginal revenue }=0$
$\displaystyle \Rightarrow 300-10x=0$
$\displaystyle \Rightarrow 10x=300$
$\displaystyle \Rightarrow x=30$
$\displaystyle \text{Thus, at output level }x=30,\text{ the marginal revenue would be zero.}$
$\\$

$\displaystyle \textbf{Question 6: } \text{The total cost function for } x \text{ units is given by }$
$\displaystyle C(x)=\sqrt{6x+5}+2500. \text{ Show that the marginal cost decreases as the output }$
$\displaystyle x \text{ increases. ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }C(x)=\sqrt{6x+5}+2500$
$\displaystyle \therefore \text{Marginal cost }(MC)=\frac{dC}{dx}$
$\displaystyle =\frac{d}{dx}\left(\sqrt{6x+5}+2500\right)$
$\displaystyle =\frac{3}{\sqrt{6x+5}}$
$\displaystyle \text{Now, }\frac{d(MC)}{dx}=\frac{d}{dx}\left(\frac{3}{\sqrt{6x+5}}\right)$
$\displaystyle =-\frac{3}{2}(6x+5)^{-3/2}\times6$
$\displaystyle =-9(6x+5)^{-3/2}$
$\displaystyle =\frac{-9}{(6x+5)^{3/2}}<0$
$\displaystyle \text{for all positive values of }x.$
$\displaystyle \text{Hence, the marginal cost falls continuously as the}$
$\displaystyle \text{output increases.}$
$\\$

$\displaystyle \textbf{Question 7: } \text{The average revenue function is given by } AR=25-\frac{x}{4}. \text{ Find total revenue}$
$\displaystyle \text{function and marginal revenue function. ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, average revenue function }(AR)=25-\frac{x}{4}$
$\displaystyle \therefore \text{Revenue function }R(x)=(AR)\cdot x$
$\displaystyle \Rightarrow R(x)=\left(25-\frac{x}{4}\right)x$
$\displaystyle =25x-\frac{x^{2}}{4}$
$\displaystyle \therefore \text{Marginal revenue }(MR)=\frac{dR}{dx}=25-\frac{2x}{4}$
$\displaystyle =25-\frac{x}{2}$
$\\$

$\displaystyle \textbf{Question 8: } \text{The demand function for a certain product is represented by the equation }$
$\displaystyle p=ax^{2}+bx+c, \text{where } x \text{ is the number of units demanded and } p \text{ is the price per unit. ISC 2023}$
$\displaystyle \text{On the basis of above information, answer the following questions.}$
$\displaystyle \text{(i) The revenue function } R(x) \text{ is}$
$\displaystyle \text{(a) } ax^{3}+bx^{2}+cx \quad \text{(b) } ax+b+\frac{c}{x} \quad \text{(c) } ax^{3}+bx^{2}+cx+d \quad \text{(d) } 2ax+b$
$\displaystyle \text{(ii) The marginal revenue } MR(x) \text{ is}$
$\displaystyle \text{(a) } a-\frac{c}{x^{2}} \quad \text{(b) } 3ax^{2}+2bx+c \quad \text{(c) } 3ax^{3}+2bx^{2}+c \quad \text{(d) } 2a$
$\displaystyle \text{(iii) The slope of the marginal revenue is}$
$\displaystyle \text{(a) } 0 \quad \text{(b) } 6ax+2b \quad \text{(c) } \frac{2c}{x^{3}} \quad \text{(d) } 9ax^{2}+4bx$
$\displaystyle \text{(iv) Values of } x, \text{ for which marginal revenue increases is}$
$\displaystyle \text{(a) } x>-\frac{b}{3a} \quad \text{(b) } x<-\frac{b}{3a} \quad \text{(c) } x=-\frac{b}{3a} \quad \text{(d) } x\leq -\frac{b}{3a}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, demand function is }p=ax^{2}+bx+c,\text{ where }x\text{ is}$
$\displaystyle \text{the number of units demanded and }p\text{ is the price per unit.}$
$\displaystyle \text{(i) (a) We know that revenue function }R=p\cdot x$
$\displaystyle \therefore R(x)=(ax^{2}+bx+c)x$
$\displaystyle =ax^{3}+bx^{2}+cx$
$\displaystyle \text{(ii) (b) We know that, marginal revenue }(MR)=\frac{dR(x)}{dx}$
$\displaystyle \therefore MR=\frac{d(ax^{3}+bx^{2}+cx)}{dx}$
$\displaystyle =3ax^{2}+2bx+c$
$\displaystyle \text{(iii) (c) We know that slope of marginal revenue }=\frac{d(MR)}{dx}$
$\displaystyle \therefore \frac{d(MR)}{dx}=\frac{d(3ax^{2}+2bx+c)}{dx}$
$\displaystyle =6ax+2b$
$\displaystyle \text{(iv) (d) For increasing marginal revenue, }\frac{d(MR)}{dx}>0$
$\displaystyle \Rightarrow 6ax+2b>0$
$\displaystyle \Rightarrow 6ax>-2b$
$\displaystyle \Rightarrow x>\frac{-2b}{6a}=\frac{-b}{3a}$
$\\$

$\displaystyle \textbf{Question 9: } \text{If the demand function is given by } p(x)=1500-2x-x^{2}, \text{ then find the}$
$\displaystyle \text{marginal revenue when } x=10. \quad \text{ISC 2023}$
$\displaystyle \text{(a) } 1160 \quad \text{(b) } 1600 \quad \text{(c) } 1100 \quad \text{(d) } 1200$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, }p(x)=1500-2x-x^{2}$
$\displaystyle \therefore R(x)=p(x)\cdot x\Rightarrow R(x)=1500x-2x^{2}-x^{3}$
$\displaystyle \therefore \text{Marginal revenue }(MR)=\frac{dR}{dx}=1500-4x-3x^{2}$
$\displaystyle \left(\frac{dR}{dx}\right)_{x=10}=1500-4(10)-3(10)^{2}$
$\displaystyle =1500-40-300$
$\displaystyle =1160$
$\\$

$\displaystyle \textbf{Question 10: } \text{The fixed cost of a product is Rs } 30000 \text{ and its variable cost per unit is}$
$\displaystyle \text{Rs } 800. \text{ If the demand function is } p(x)=4500-100x, \text{ then find the break-even values.}$
$\displaystyle \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, fixed cost}=\text{Rs }30000$
$\displaystyle \text{Variable cost}=\text{Rs }800\text{ per unit}$
$\displaystyle \therefore \text{Cost function, }C(x)=800x+30000$
$\displaystyle \text{and Demand function, }p(x)=4500-100x$
$\displaystyle \therefore \text{Revenue function, }R(x)=p(x)\cdot x$
$\displaystyle =(4500-100x)x$
$\displaystyle =4500x-100x^{2}$
$\displaystyle \text{For break-even point,}$
$\displaystyle R(x)=C(x)$
$\displaystyle \Rightarrow 4500x-100x^{2}=800x+30000$
$\displaystyle \Rightarrow 100x^{2}-3700x+30000=0$
$\displaystyle \Rightarrow x^{2}-37x+300=0$
$\displaystyle \Rightarrow x^{2}-25x-12x+300=0$
$\displaystyle \Rightarrow (x-25)(x-12)=0$
$\displaystyle \Rightarrow x=25,12$
$\displaystyle \text{Break-even values}=25,12$
$\\$

$\displaystyle \textbf{Question 11: } \text{The cost function } C(x)=3x^{2}-6x+5. \text{ Find the average cost when }$
$\displaystyle x=2. \ \ \ \ \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }C(x)=3x^{2}-6x+5$
$\displaystyle \therefore \text{Average cost }(AC)=\frac{C(x)}{x}=\frac{3x^{2}-6x+5}{x}=3x-6+\frac{5}{x}$
$\displaystyle \text{Average cost at }x=2\text{ is}$
$\displaystyle 3(2)-6+\frac{5}{2}=6-6+\frac{5}{2}=\frac{5}{2}$
$\\$

$\displaystyle \textbf{Question 12: } \text{The selling price of a commodity is fixed at Rs } 60 \\ \text{ and its cost function is } C(x)=35x+250.$
$\displaystyle \text{(i) Determine the profit function. (ii) Find the break-even points. ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, price of commodity = Rs }60$
$\displaystyle R(x)=px=60x$
$\displaystyle C(x)=35x+250$
$\displaystyle \text{(i) Profit function }P(x)=R(x)-C(x)$
$\displaystyle \Rightarrow P(x)=60x-35x-250=25x-250$
$\displaystyle \text{(ii) At break-even point, }R(x)=C(x)$
$\displaystyle \Rightarrow 60x=35x+250$
$\displaystyle \Rightarrow 25x=250\Rightarrow x=10$
$\displaystyle \text{Hence, the break-even point is at }x=10$
$\\$

$\displaystyle \textbf{Question 13: } \text{The revenue function is given by } R(x)=100x-x^{2}-x^{3}. \text{ Find}$
$\displaystyle \text{(i) the demand function. (ii) marginal revenue function. ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }R(x)=100x-x^{2}-x^{3}$
$\displaystyle \text{(i) Demand function, }p=\frac{R(x)}{x}$
$\displaystyle =\frac{100x-x^{2}-x^{3}}{x}=100-x-x^{2}$
$\displaystyle \text{(ii) Marginal revenue function }(MR)=\frac{d(R(x))}{dx}$
$\displaystyle =\frac{d(100x-x^{2}-x^{3})}{dx}$
$\displaystyle =100-2x-3x^{2}$
$\\$

$\displaystyle \textbf{Question 14: } \text{The total cost function of a firm is given by }$
$\displaystyle C(x)=\frac{1}{3}x^{3}-5x^{2}+30x-15,\text{where the selling price per unit is given as Rs } 6.$
$\displaystyle \text{Find for what value of } x \text{ will the profit be maximum. ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }C(x)=\frac{1}{3}x^{3}-5x^{2}+30x-15$
$\displaystyle R(x)=px$
$\displaystyle \therefore R(x)=6x\quad [\because p=6]$
$\displaystyle \text{Profit function}$
$\displaystyle P(x)=R(x)-C(x)$
$\displaystyle P(x)=6x-\frac{1}{3}x^{3}+5x^{2}-30x+15$
$\displaystyle P(x)=-\frac{1}{3}x^{3}+5x^{2}-24x+15$
$\displaystyle \frac{dP(x)}{dx}=-x^{2}+10x-24$
$\displaystyle \text{For maxima or minima, }\frac{dP(x)}{dx}=0$
$\displaystyle \Rightarrow -x^{2}+10x-24=0$
$\displaystyle \Rightarrow x^{2}-10x+24=0$
$\displaystyle \Rightarrow (x-6)(x-4)=0$
$\displaystyle \Rightarrow x=6,4$
$\displaystyle \frac{d^{2}P(x)}{dx^{2}}=-2x+10$
$\displaystyle \left(\frac{d^{2}P(x)}{dx^{2}}\right)_{x=6}=-12+10=-2<0$
$\displaystyle \left(\frac{d^{2}P(x)}{dx^{2}}\right)_{x=4}=-8+10=2>0$
$\displaystyle \text{Hence, profit is maximum when }x=6$
$\\$

$\displaystyle \textbf{Question 15: } \text{A company produces a commodity with Rs } 24000 \text{ as fixed cost.}$
$\displaystyle \text{The variable cost, estimated to be } 25\% \text{ of total revenue received on selling the product,}$
$\displaystyle \text{is at the rate of Rs 8 per unit. Find the break-even point. ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, fixed cost of the commodity = Rs }24000$
$\displaystyle \text{Let the number of units produced by company be }x\text{ and}$
$\displaystyle p\text{ be the price per unit.}$
$\displaystyle \text{Then, total revenue }R=xp=\text{Rs }8x$
$\displaystyle \text{Since, the variable cost is }25\%\text{ of the total revenue.}$
$\displaystyle \text{Thus, variable cost}=25\%\text{ of }8x$
$\displaystyle =\frac{25}{100}\times8x=\frac{1}{4}\times8x=\text{Rs }2x$
$\displaystyle \text{Total cost }C(x)=\text{Fixed cost of the commodity}$
$\displaystyle +\text{Variable cost of the commodity}$
$\displaystyle =24000+2x$
$\displaystyle \text{For break-even point, }R(x)=C(x)$
$\displaystyle \Rightarrow 8x=24000+2x$
$\displaystyle \Rightarrow 8x-2x=24000$
$\displaystyle \Rightarrow 6x=24000$
$\displaystyle \therefore x=\frac{24000}{6}=4000$
$\displaystyle \text{Hence, the break-even point is }4000.$
$\\$

$\displaystyle \textbf{Question 16: } \text{The cost function of a product is given by }$
$\displaystyle C(x)=\frac{x^{3}}{3}-45x^{2}-900x+36, \text{ where } x \text{ is the number of units produced. How many }$
$\displaystyle \text{ units should be produced to minimise the marginal cost? ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, cost function is}$
$\displaystyle C(x)=\frac{x^{3}}{3}-45x^{2}-900x+36$
$\displaystyle \therefore \text{Marginal Cost, }MC=\frac{dC}{dx}$
$\displaystyle =\frac{d}{dx}\left(\frac{x^{3}}{3}-45x^{2}-900x+36\right)$
$\displaystyle =\frac{3x^{2}}{3}-90x-900$
$\displaystyle \Rightarrow MC=x^{2}-90x-900\quad ...(i)$
$\displaystyle \text{On differentiating Eq. (i) w.r.t. }x,\text{ we get}$
$\displaystyle \frac{d(MC)}{dx}=2x-90\text{ and }\frac{d^{2}(MC)}{dx^{2}}=2$
$\displaystyle \text{For minimum marginal cost, put }\frac{d(MC)}{dx}=0$
$\displaystyle \Rightarrow 2x-90=0\Rightarrow x=45$
$\displaystyle \text{At }x=45,\;\frac{d^{2}(MC)}{dx^{2}}=2>0$
$\displaystyle \text{Hence, it produces }45\text{ units, to minimise the marginal cost.}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Given, the total cost function for } x \text{ units of a commodity as }$
$\displaystyle C(x)=\frac{1}{3}x^{3}+3x^{2}-16x+2. \text{ Find (i) marginal cost function (ii) average cost function. ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }C(x)=\frac{1}{3}x^{3}+3x^{2}-16x+2$
$\displaystyle \text{(i) Marginal Cost }(MC)=\frac{dC}{dx}$
$\displaystyle \therefore MC=\frac{d}{dx}\left(\frac{1}{3}x^{3}+3x^{2}-16x+2\right)$
$\displaystyle MC=x^{2}+6x-16$
$\displaystyle \text{(ii) Average Cost }(AC)=\frac{C}{x}$
$\displaystyle AC=\frac{\frac{1}{3}x^{3}+3x^{2}-16x+2}{x}$
$\displaystyle \Rightarrow AC=\frac{x^{2}}{3}+3x-16+\frac{2}{x}$
$\\$

$\displaystyle \textbf{Question 18: } \text{The average cost function associated with producing and marketing } x$
$\displaystyle \text{ units of an item is given by} AC=2x-11+\frac{50}{x}. \text{ Find the range of values of the output }$
$\displaystyle x, \text{ for which } AC \text{ is increasing. ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, Average cost }AC=2x-11+\frac{50}{x}$
$\displaystyle \text{On differentiating w.r.t. }x,\text{ we get}$
$\displaystyle \frac{d}{dx}(AC)=2-\frac{50}{x^{2}}$
$\displaystyle \text{For average cost is increasing}$
$\displaystyle \frac{d}{dx}(AC)>0$
$\displaystyle \therefore 2-\frac{50}{x^{2}}>0$
$\displaystyle \Rightarrow 2x^{2}-50>0\quad [\because x^{2}>0]$
$\displaystyle \Rightarrow x^{2}>25$
$\displaystyle \Rightarrow x>5$
$\displaystyle \text{Hence, the average cost increases, if the output }x>5.$
$\\$

$\displaystyle \textbf{Question 19: } \text{A product can be manufactured at a total cost }$
$\displaystyle C(x)=\frac{x^{2}}{100}+100x+40, \text{ where } x \text{ is the number of units produced. }$
$\displaystyle \text{The price at which each unit can be sold is } P=\left(200-\frac{x}{400}\right).$
$\displaystyle \text{Determine the production level } x \text{ at which the profit is maximum. What is }$
$\displaystyle \text{the price per unit and total profit } \text{at the level of production? ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, total cost }C(x)=\frac{x^{2}}{100}+100x+40$
$\displaystyle \text{Price of each unit can be sold }P=200-\frac{x}{400}$
$\displaystyle \text{Revenue }R(x)=P\cdot x=200x-\frac{x^{2}}{400}$
$\displaystyle \text{Profit, }P(x)=R(x)-C(x)$
$\displaystyle =200x-\frac{x^{2}}{400}-\frac{x^{2}}{100}-100x-40$
$\displaystyle =200x-\frac{x^{2}}{400}-\frac{4x^{2}}{400}-100x-40$
$\displaystyle =100x-\frac{5x^{2}}{400}-40$
$\displaystyle =100x-\frac{x^{2}}{80}-40$
$\displaystyle \text{On differentiating w.r.t. }x,\text{ we get}$
$\displaystyle P'(x)=100-\frac{2x}{80}=100-\frac{x}{40}$
$\displaystyle \text{For maxima or minima put }P'(x)=0$
$\displaystyle \Rightarrow 100-\frac{x}{40}=0\Rightarrow x=4000$
$\displaystyle \text{Now, }P''(x)=-\frac{1}{40}<0$
$\displaystyle \text{Hence, }P(x)\text{ is maximum when }x=4000$
$\displaystyle \text{Total profit at }x=4000,$
$\displaystyle P(4000)=100(4000)-\frac{(4000)^{2}}{80}-40$
$\displaystyle =400000-200000-40=199960$
$\displaystyle \text{Total profit = Rs }199960$
$\\$

$\displaystyle \textbf{Question 20: } \text{The demand for a certain product is represented by the equation }$
$\displaystyle p=500+25x-\frac{x^{2}}{3} \text{in rupees where } x \text{ is the number of units and } p$
$\displaystyle \text{ is the price per unit. Find (i) marginal revenue function}$
$\displaystyle \text{(ii) the marginal revenue when } 10 \text{ units are sold. ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ Given, }p=500+25x-\frac{x^{2}}{3}$
$\displaystyle \text{Clearly, total revenue}$
$\displaystyle R(x)=px=\left(500+25x-\frac{x^{2}}{3}\right)x$
$\displaystyle =500x+25x^{2}-\frac{x^{3}}{3}$
$\displaystyle \text{(i) We know that, marginal revenue }=\frac{dR(x)}{dx}$
$\displaystyle \therefore \frac{dR(x)}{dx}=500+50x-x^{2}$
$\displaystyle \text{(ii) Marginal revenue at }x=10,\text{ is given by}$
$\displaystyle \left(\frac{dR}{dx}\right)_{x=10}=500+50(10)-(10)^{2}$
$\displaystyle =500+500-100=900$
$\\$

$\displaystyle \textbf{Question 21: } \text{The demand function is } x=\frac{24-2p}{3}, \text{ where } x \text{ is the number of units demanded}$
$\displaystyle \text{and } p \text{ is the price per unit. (i) Find the revenue function } R \text{ in terms of } p. \text{ (ii) Find the price}$
$\displaystyle \text{and the number of units demanded for which the revenue is maximum. ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, demand function, }x=\frac{24-2p}{3}$
$\displaystyle \text{Revenue function, }R(p)=\text{Demand function}\times\text{Per unit price}$
$\displaystyle =\frac{24-2p}{3}\cdot p=\frac{(24-2p)p}{3}$
$\displaystyle \text{(ii) For revenue to be maximum, put }\frac{dR}{dp}=0$
$\displaystyle \therefore \frac{d}{dp}\left(\frac{24p-2p^{2}}{3}\right)=0$
$\displaystyle \Rightarrow \frac{1}{3}(24-4p)=0$
$\displaystyle \Rightarrow 24-4p=0$
$\displaystyle \Rightarrow p=6$
$\displaystyle \text{Now, }\frac{d^{2}R}{dp^{2}}=-\frac{4}{3}<0$
$\displaystyle \therefore \text{Revenue is maximum when }p=6$
$\displaystyle \text{Hence, number of units demanded for maximum revenue is}$
$\displaystyle x=\frac{24-2(6)}{3}=\frac{12}{3}=4\text{ units.}$
$\\$

$\displaystyle \textbf{Question 22: } \text{The average cost function } AC \text{ for a commodity is given by }$
$\displaystyle AC=x+5+\frac{36}{x} \text{in terms of output } x, \text{ find the (i) marginal cost as the function of } x.$
$\displaystyle \text{ (ii) output for which } AC \text{ increases. ISC 2015, 10}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, the average cost function of the commodity is}$
$\displaystyle (AC)=x+5+\frac{36}{x}$
$\displaystyle \text{We know that, Average cost }(AC)=\frac{\text{Cost function}}{x}$
$\displaystyle \Rightarrow C=x(AC)$
$\displaystyle \Rightarrow C=x\left(x+5+\frac{36}{x}\right)$
$\displaystyle \Rightarrow C=x^{2}+5x+36$
$\displaystyle \text{(i) Marginal Cost function}$
$\displaystyle (MC)=\frac{dC}{dx}=\frac{d}{dx}(x^{2}+5x+36)$
$\displaystyle \Rightarrow MC=2x+5$
$\displaystyle \text{(ii) Condition for increasing of AC,}$
$\displaystyle \frac{d}{dx}(AC)>0$
$\displaystyle \Rightarrow \frac{d}{dx}\left(x+5+\frac{36}{x}\right)>0$
$\displaystyle \Rightarrow 1-\frac{36}{x^{2}}>0$
$\displaystyle \Rightarrow \frac{x^{2}-36}{x^{2}}>0$
$\displaystyle \Rightarrow (x^{2}-6^{2})>0$
$\displaystyle \Rightarrow (x+6)(x-6)>0$
$\displaystyle \Rightarrow x<-6\text{ or }x>6$
$\displaystyle \text{So, the average cost of the commodity increases, when the output }x>6.$
$\\$

$\displaystyle \textbf{Question 23: } \text{A firm has the cost function, } C=\frac{x^{3}}{3}-7x^{2}+111x+50$
$\displaystyle \text{and demand function, } x=100-p.$
$\displaystyle \text{(i) Write the total revenue in terms of } x. \text{ (ii) Formulate the total profit function } P \text{ in terms of } x.$
$\displaystyle \text{(iii) Find the profit maximising level of output } x. \text{ ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, cost function, }C=\frac{x^{3}}{3}-7x^{2}+11x+50$
$\displaystyle \text{and demand function, }x=100-P$
$\displaystyle \text{We know that,}$
$\displaystyle \text{(i) Revenue function, }R=xP$
$\displaystyle \text{From equation of demand function, }x=100-P$
$\displaystyle \Rightarrow P=100-x$
$\displaystyle \text{Now, we put the value of }P=100-x\text{ in equation of}$
$\displaystyle \text{revenue function, we get}$
$\displaystyle R=x(100-x)\Rightarrow R=100x-x^{2}$
$\displaystyle \text{(ii) Also, profit function }P(x)=\text{Revenue}-\text{Cost}$
$\displaystyle \Rightarrow P(x)=100x-x^{2}-\left(\frac{x^{3}}{3}-7x^{2}+11x+50\right)$
$\displaystyle \Rightarrow P(x)=100x-x^{2}-\frac{x^{3}}{3}+7x^{2}-11x-50$
$\displaystyle \Rightarrow P(x)=-\frac{x^{3}}{3}+6x^{2}+89x-50$
$\displaystyle \text{(iii) For the profit maximising level of output }x,\ \frac{dP}{dx}=0$
$\displaystyle \Rightarrow \frac{d}{dx}\left(-\frac{x^{3}}{3}+6x^{2}+89x-50\right)=0$
$\displaystyle \Rightarrow -x^{2}+12x+89=0$
$\displaystyle \Rightarrow x^{2}-12x-89=0$
$\displaystyle \Rightarrow x=\frac{12\pm\sqrt{144+356}}{2}=\frac{12\pm\sqrt{500}}{2}$
$\displaystyle =\frac{12\pm10\sqrt{5}}{2}=6\pm5\sqrt{5}$
$\displaystyle \text{Now, }\frac{d^{2}P}{dx^{2}}=\frac{d}{dx}(-x^{2}+12x+89)=-2x+12$
$\displaystyle \text{At }x=6-5\sqrt{5},\ \frac{d^{2}P}{dx^{2}}>0\ (\text{minimum})$
$\displaystyle \text{At }x=6+5\sqrt{5},\ \frac{d^{2}P}{dx^{2}}<0\ (\text{maximum})$
$\displaystyle \text{Hence, profit is maximum at }x=6+5\sqrt{5}.$
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